The univalent Bloch-Landau constant, harmonic symmetry and ...

More documents

Recommendations

Info

the extremal configuration will be harmonically symmetric at infinity. As noted in the introduction, this latter observation was first made by Lavrentiev [15]. Wework withdomains Ω = Ω z0 ,R asshown inFigure4, where R > 3, |z 0 | < R 3 , and the arc γ z0 is chosen so that Ω z0 ,R is harmonically symmetric in γ z0 with respect to 0. If g is a conformal map of the unit 7 γ z0 Ω z0 ,R z 0 −8 1 z 0 R 3 γ z0 Figure 4. A domain Ω z0 ,R disk D onto such a domain Ω z0 ,R, with g(0) = 0, then f(z) = z 3 √ g(z 3 ) z 3 is a conformal map of D onto a domain U w,R as shown in Figure 5. The arcs that appear are all harmonically symmetric, and thus the w U w,R −2 1 R w Figure 5. A domain U w,R conformal mapping of the unit disk onto U w,R is a good candidate for having a relatively large derivative at the origin. This derivative is |f ′ (0)| = 3√ |g ′ (0)|. In order that this provide a useful estimate of the Bloch-Landau constant, we need to arrange for U w,R to have inradius 1. We leave this aside for the moment and show how to use Fedorov’s results on capacity to compute |f ′ (0)| for given z 0 and R. We write k for the Koebe mapping k(z) = z/(1 −z) 2 of the unit disk onto the plane slit along the negative real axis from minus infinity to −1/4.
8 Proposition 1. We write f for a conformal map of the unit disk D onto U w,R for which f(0) = 0. Then, with z 0 = w 3 , (3.3) |f ′ (0)| = 1 3 √|ψ(z 0 )−ψ(1)|cap ( E(α,c) ) R where 1 (3.4) ψ(z) = − k(z/R 3 ) and (3.5) e iα = ψ(z 0)−ψ(1) |ψ(z 0 )−ψ(1)| , Proof. The map (3.6) φ(z) = ψ(z)−ψ(1) |ψ(z 0 )−ψ(1)| , c = ψ(−8)−ψ(1) |ψ(z 0 )−ψ(1)| . where ψ isgiven by (3.4), maps Ω z0 ,R ontothe complement of Fedorov’s continuum E(α,c) with α and c given by (3.5). The harmonic symmetry of arcs is preserved because each mapping extends continuously to all internal boundary arcs and because the domains involved are all symmetric with respect to the real axis. If h is the mapping of the complement of E(α,c) onto the complement of the unit disk mentioned in (3.2), then a suitable mapping g of Ω z0 ,R onto the unit disk, with g(0) = 0, is given by (3.7) g(z) = 1 h ( φ(z) ). Now g ′ (0) can be computed explicitly, in terms of z 0 , R and h ′ (∞), for example by computing the power series for g. One obtains (3.8) |g ′ (0)| = |ψ(z 0)−ψ(1)| . R 3 h ′ (∞) The value of h ′ (∞) is 1/cap ( E(α,c) ) and is given explicitly, in its turn, by Fedorov’s result (3.1). □ 3.3. The choice of w and R. We know how to build, for any w and R, a conformal map f of the unit disk onto U w,R and have a formula for its derivative at 0. In order that this provide an upper bound for the univalent Bloch-Landau constant, we must choose w and R in such a way that • The domain U w,R has inradius one, • The derivative |f ′ (0)| is as big as possible. Proposition 2. We suppose that R is fixed with 3 < R < 4.5. We set (3.9) P 2 = (R−1)e i(π 3 −α) where α = arcsin [ 1/(R−1) ] ,
Page 1 and 2: The univalent Bloch-Landau constant
Page 3 and 4: 2 Figure 1. The domain D 0 and the
Page 5 and 6: 4 There is an admissible family of
Page 7: 6 harmonic symmetry and the Pólya-
Page 11 and 12: 10 For the actual picture of the do
Page 13 and 14: 12 Morera’s Theorem is applicable
Page 15 and 16: 14 the boundary of the domain D \

The univalent Bloch-Landau constant, harmonic symmetry and ...

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?